52639 + Interview Questions in Secondary School in Chemistry

1.	निम्नलिखित उपसहसंयोजन सत्ता में धातुओं के ऑक्सीकरण अंक का उल्लेख कीजिए- [tex] (i) [Co(H_2O)(CN)(en)_2]^{2+} (iii) [PtCl_4]^{2–} (v) [Cr(NH_3)_3Cl_3](ii) [CoBr_2(en)_2]+ (iv) K_3[Fe(CN)_6][/tex]
Answer» ANSWER: I)Co-+3 iii)Pt-+2 iv)Fe-+3

1.

निम्नलिखित उपसहसंयोजन सत्ता में धातुओं के ऑक्सीकरण अंक का उल्लेख कीजिए- [tex] (i) [Co(H_2O)(CN)(en)_2]^{2+} (iii) [PtCl_4]^{2–} (v) [Cr(NH_3)_3Cl_3](ii) [CoBr_2(en)_2]+ (iv) K_3[Fe(CN)_6][/tex]

Answer»

ANSWER:

I)Co-+3

iii)Pt-+2

iv)Fe-+3

Discussion

2.	विलयन तथा विलयन का 1:1 मोलर अनुपात में मिश्रण Fe2+ आयन का परीक्षण देता है परंतु CuSO4 व जलीय अमोनिया का 1:4 मोलर अनुपात में मिश्रण Cu2+आयनों का परीक्षण नहीं देता। समझाइए क्यों?
Answer» ANSWER: PLEASE GIVE in english Explanation:

2.

विलयन तथा विलयन का 1:1 मोलर अनुपात में मिश्रण Fe2+ आयन का परीक्षण देता है परंतु CuSO4 व जलीय अमोनिया का 1:4 मोलर अनुपात में मिश्रण Cu2+आयनों का परीक्षण नहीं देता। समझाइए क्यों?

Answer»

ANSWER:

PLEASE GIVE in english

Explanation:

Discussion

3.	Is the solid state of CO2
Answer» Answer: Carbon dioxide is colorless. ... At 1 atmosphere (NEAR mean sea level pressure), the gas DEPOSITS DIRECTLY to a SOLID at TEMPERATURES below −78.5 °C (−109.3 °F; 194.7 K) and the solid sublimes directly to a gas above −78.5 °C. In its solid state, carbon dioxide is commonly called dry ice. I HOPE THIS MAY HELP U.. #GUNDI.. FOLLOW ME MARK AS BRAINLIST..✌✌

3.

Is the solid state of CO2

Answer»

Answer:

Carbon dioxide is colorless. ... At 1 atmosphere (NEAR mean sea level pressure), the gas DEPOSITS DIRECTLY to a SOLID at TEMPERATURES below −78.5 °C (−109.3 °F; 194.7 K) and the solid sublimes directly to a gas above −78.5 °C. In its solid state, carbon dioxide is commonly called dry ice.

I HOPE THIS MAY HELP U..

#GUNDI..

FOLLOW ME

MARK AS BRAINLIST..✌✌

Discussion

4.	If volume percentage of N2 is 78.than weight percentage of the same is
Answer» Answer: A particle has a displacement of 40m along OX in the first 5S and 30M perpendicular to OX in the next 5s. The magnitude of average velocity of the particle during the time internal of 10S is a) 6 ms–1 B) 7 ms–1 C) 8 ms–1 d) 5 ms–1 [

4.

If volume percentage of N2 is 78.than weight percentage of the same is

Answer»

Answer:

A particle has a displacement of 40m along OX in the first 5S and 30M perpendicular to OX in

the next 5s. The magnitude of average velocity of the particle during the time internal of 10S is

a) 6 ms–1 B) 7 ms–1 C) 8 ms–1 d) 5 ms–1 [

Discussion

5.	How we get the ratio in 16:1:2
Answer» ADD these no. 16:1:2 you GET = 19/1 19:1 ANS..... i HOPE i HELPED you

5.

How we get the ratio in 16:1:2

Answer»

ADD these no. 16:1:2

you GET = 19/1

19:1 ANS.....

i HOPE i HELPED you

Discussion

6.	1. Which of the following represents largest number of particles?a. atoms in 1 mole of CH4b. atoms in 0.5 mole of SO,c. atoms in 0.5 mole of CO4d. atoms in 1 mole of CO
Answer» ANSWER: a)ATOMS in 1 MOLE of CH4

6.

1. Which of the following represents largest number of particles?a. atoms in 1 mole of CH4b. atoms in 0.5 mole of SO,c. atoms in 0.5 mole of CO4d. atoms in 1 mole of CO

Answer»

ANSWER:

a)ATOMS in 1 MOLE of CH4

Discussion

7.	C3h8+5o2=3co2+4h20 find ∆ng
Answer» Answer: $\displaystyle \sf \longrightarrow \Delta \ n_g=-3 \\$ Explanation: Given : $\displaystyle \sf C_3H_8+5 \ O_2( g) \rightleftharpoons 3 \ CO_2 (g)+ 4 \ H_2O(l) \\ \\$ We are asked to FIND VALUE of Δ n g : Since we know Δ n g is DIFFERENCE of sum of product moles and sum of reactant moles of GASES. We will not consider liquid one! $\displaystyle \sf \longrightarrow \Delta \ n_g=(3)-(1+5) \\ \\$ $\displaystyle \sf \longrightarrow \Delta \ n_g=3-6 \\ \\$ $\displaystyle \sf \longrightarrow \Delta \ n_g=-3 \\ \\$ Therefore we get required answer. Read more : Give the relation between K_p and K_c for the for the reaction 2 NaCl ( g ) ⇄ 2 Na ( g ) + Cl₂ ( g ). brainly.in/question/16770768

7.

C3h8+5o2=3co2+4h20 find ∆ng

Answer»

Answer:

$\displaystyle \sf \longrightarrow \Delta \ n_g=-3 \\$

Explanation:

Given :

$\displaystyle \sf C_3H_8+5 \ O_2( g) \rightleftharpoons 3 \ CO_2 (g)+ 4 \ H_2O(l) \\ \\$

We are asked to FIND VALUE of Δ n g :

Since we know Δ n g is DIFFERENCE of sum of product moles and sum of reactant moles of GASES. We will not consider liquid one!

$\displaystyle \sf \longrightarrow \Delta \ n_g=(3)-(1+5) \\ \\$

$\displaystyle \sf \longrightarrow \Delta \ n_g=3-6 \\ \\$

$\displaystyle \sf \longrightarrow \Delta \ n_g=-3 \\ \\$

Therefore we get required answer.

Read more :

Give the relation between K_p and K_c for the for the reaction 2 NaCl ( g ) ⇄ 2 Na ( g ) + Cl₂ ( g ).

brainly.in/question/16770768

Discussion

8.	22 g of Ar find the mole
Answer» Answer: 0.55 MOLE...........

8.

22 g of Ar find the mole

Answer»

Answer:

0.55 MOLE...........

Discussion

9.	(a) Giving reason identify the susanca ore and the substance reduced in each of the following chemical reactions :(i) H2 + CuO --> Cu + H2O(ii) H2S + SO2 --> 2S + 2H2O(b) Name the term used to represent a reaction in which ondation and reduction takepace simultaneously
Answer» Answer : ★ Oxidation : The process of adding OXYGEN to a substance is called oxidation. ★ Reduction : The process of losing oxygen from a substance is called reduction. $\hrulefill$ (a) (i) Substance oxidised : $\bold{H_2}$ $\bold{H_2}$ is oxidised here since it gains oxygen to become $\bold{H_2O}$ . Substance reduced : $\bold{CuO}$ $\bold{CuO}$ is reduced here since it LOSES oxygen to become $\bold{Cu}$ $\hrulefill$ (ii) Substance oxidised : $\bold{H_2S}$ $\bold{H_2S}$ is oxidised here since it gains oxygen in the given REACTION. Substance reduced : $\bold{SO_2}$ $\bold{SO_2}$ is being reduced here as it loses oxygen in the given reaction to become $\bold{S}$ $\hrulefill$ (b) A reaction in which oxidation & reduction takes PLACE SIMULTANEOUSLY is called a redox reaction.

9.

(a) Giving reason identify the susanca ore and the substance reduced in each of the following chemical reactions :(i) H2 + CuO --> Cu + H2O(ii) H2S + SO2 --> 2S + 2H2O(b) Name the term used to represent a reaction in which ondation and reduction takepace simultaneously

Answer»

Answer :

★ Oxidation : The process of adding OXYGEN to a substance is called oxidation.

★ Reduction : The process of losing oxygen from a substance is called reduction.

$\hrulefill$

(a) (i) Substance oxidised : $\bold{H_2}$

$\bold{H_2}$ is oxidised here since it gains oxygen to become $\bold{H_2O}$ .

Substance reduced : $\bold{CuO}$

$\bold{CuO}$ is reduced here since it LOSES oxygen to become $\bold{Cu}$

$\hrulefill$

(ii) Substance oxidised : $\bold{H_2S}$

$\bold{H_2S}$ is oxidised here since it gains oxygen in the given REACTION.

Substance reduced : $\bold{SO_2}$

$\bold{SO_2}$ is being reduced here as it loses oxygen in the given reaction to become $\bold{S}$

$\hrulefill$

(b) A reaction in which oxidation & reduction takes PLACE SIMULTANEOUSLY is called a redox reaction.

Discussion

10.	Mass of total positive charge present in an atom is 16533 times to that of mass of electron.find the atomic number of this element.
Answer» 2, 8 , 9 , 2 2 , 8 , 18 , 1 2, 8 , 18 , 2 2 , 8 , 8 2 Answer : 3 Solution : We know that mass of a proton is 1837 times that of an electron. ∴ Atomic number of element =551101837=30 ∴ Electronic CONFIGURATION of the element is 2 , 8 , 18 , 2

10.

Mass of total positive charge present in an atom is 16533 times to that of mass of electron.find the atomic number of this element.

Answer»

2, 8 , 9 , 2

2 , 8 , 18 , 1

2, 8 , 18 , 2

2 , 8 , 8 2

Answer :

3

Solution :

We know that mass of a proton is 1837 times that of an electron.

∴ Atomic number of element =551101837=30

∴ Electronic CONFIGURATION of the element is 2 , 8 , 18 , 2

Discussion

11.	Evaporation is a surface phenomenan
Answer» ANSWER: YES evaporation is a surface PHENOMENON since higher surface AREA INCREASES ,evaporation increases.

11.

Evaporation is a surface phenomenan

Answer»

ANSWER:

YES evaporation is a surface PHENOMENON

since higher surface AREA INCREASES ,evaporation increases.

Discussion

12.	CuSO4.5H2O belongs to-triclinic system or cubic system
Answer» Answer: A triclinic crystal is shaped almost LIKE a rectangle. It is slightly bent, though. Copper SULFATE makes triclinic crystals. It is the least symmetric (meaning that both SIDES are not very similar). . In the triclinic system, the crystal is described by vectors of UNEQUAL LENGTH, as in the orthorhombic system.

12.

CuSO4.5H2O belongs to-triclinic system or cubic system

Answer»

Answer:

A triclinic crystal is shaped almost LIKE a rectangle. It is slightly bent, though. Copper SULFATE makes triclinic crystals. It is the least symmetric (meaning that both SIDES are not very similar).

. In the triclinic system, the crystal is described by vectors of UNEQUAL LENGTH, as in the orthorhombic system.

Discussion

13.	How many moles are in 1.25 x 1025 particles of NaCl?a. 20.8 mol NaClb. 7.53 x 1048 moles NaClc. 7.53 mol NaCld. 2.08 mol NaClplease answer
Answer» ANSWER: The answer will be :- OPTION a. 20.8 MOL NACL.

13.

How many moles are in 1.25 x 1025 particles of NaCl?a. 20.8 mol NaClb. 7.53 x 1048 moles NaClc. 7.53 mol NaCld. 2.08 mol NaClplease answer

Answer»

ANSWER:

The answer will be :-

OPTION a. 20.8 MOL NACL.

Discussion

14.	Find density ofIceWaterfreonDiamondAluminiumWoodoilLeadIodine.
Answer» ANSWER: Ice=0.92 Aluminium=2.7 Oil=0.87 Lead=11.5

14.

Find density ofIceWaterfreonDiamondAluminiumWoodoilLeadIodine.

Answer»

ANSWER:

Ice=0.92

Aluminium=2.7

Oil=0.87

Lead=11.5

Discussion

15.	Calculate the amount of carbon dioxide produce in the combustion of 24gm of carbon
Answer» ANSWER: AMOUNT of CARBO dioxide is ALSO 48 GM

15.

Calculate the amount of carbon dioxide produce in the combustion of 24gm of carbon

Answer»

ANSWER:

AMOUNT of CARBO dioxide is ALSO 48 GM

Discussion

16.	21+31=52 . write in scientific notion.....
Answer»

Discussion

17.	Examples of slow and fast changes five of them
Answer» The changes which take place in a long period of time are called slow changes whereas that changes which take place in a short period of time are called FAST changes. Examples: (a) Rusting of iron, formation of DAY and night, ripening of fruits, GROWING of trees are slow changes. (b) Burning of paper, stretching of rubber band, blowing of balloons, bursting of crackers are fast changes. hope this helps you MATE.. MARK ME BRAINLIST..!!

17.

Examples of slow and fast changes five of them

Answer»

The changes which take place in a long period of time are called slow changes whereas that changes which take place in a short period of time are called FAST changes.

Examples:

(a) Rusting of iron, formation of DAY and night, ripening of fruits, GROWING of trees are slow changes.

(b) Burning of paper, stretching of rubber band, blowing of balloons, bursting of crackers are fast changes.

hope this helps you MATE..

MARK ME BRAINLIST..!!

Discussion

18.	इसका प्रमाण दीजिए कि तथा आयनन समावय हैं।
Answer» ANSWER: it is TEX is the answer

18.

इसका प्रमाण दीजिए कि तथा आयनन समावय हैं।

Answer»

ANSWER:

it is TEX is the answer

Discussion

19.	निम्नलिखित संकुलों द्वारा प्रदर्शित समावयवता का प्रकार बतलाइए तथा इन समावयवों की संरचनाएं बनाए। (i) K[Cr(H2O)2(C2O4)2] (ii) K[Co(en)3]Cl3 (iii) [Co(NH3)5(NO2)] (NO3)2 (iv) [Pt(NH3)(H2O)Cl2]
Answer» Explanation: HYE..... (iv) [Pt(NH3)(H2O)Cl2] Is CORRECT ANSWER Hope it's help Mark as brainliest

19.

निम्नलिखित संकुलों द्वारा प्रदर्शित समावयवता का प्रकार बतलाइए तथा इन समावयवों की संरचनाएं बनाए। (i) K[Cr(H2O)2(C2O4)2] (ii) K[Co(en)3]Cl3 (iii) [Co(NH3)5(NO2)] (NO3)2 (iv) [Pt(NH3)(H2O)Cl2]

Answer»

Explanation:

HYE.....

(iv) [Pt(NH3)(H2O)Cl2]

Is CORRECT ANSWER

Hope it's help

Mark as brainliest

Discussion

20.	निम्नलिखित उपसहसंयोजन यौगिकों के सूत्र लिखिए-(i) टेट्राऐम्मीनडाइएक्वाकोबाल्ट (III) क्लोराइड (ii) पोटेशियम टेट्रासायनिडोनिकैलेट (II) (iii) ट्रिस(एथेन-1, 2-डाइऐमीन)क्रोमियम (III) क्लोराइड (iv) ऐम्मीनब्रोमिडोक्लोरिडॉनाइट्रिटो-N-प्लैटिनेट (II) (v) डाइक्लोरोबिस(एथेन-1, 2-डाइऐमीन)प्लैटिनम (IV) नाइट्रेट(vi) आयरन(III)हेक्सासायनिडोफेरेट(II)
Answer» OPTION no 2 is CORRECT.

20.

निम्नलिखित उपसहसंयोजन यौगिकों के सूत्र लिखिए-(i) टेट्राऐम्मीनडाइएक्वाकोबाल्ट (III) क्लोराइड (ii) पोटेशियम टेट्रासायनिडोनिकैलेट (II) (iii) ट्रिस(एथेन-1, 2-डाइऐमीन)क्रोमियम (III) क्लोराइड (iv) ऐम्मीनब्रोमिडोक्लोरिडॉनाइट्रिटो-N-प्लैटिनेट (II) (v) डाइक्लोरोबिस(एथेन-1, 2-डाइऐमीन)प्लैटिनम (IV) नाइट्रेट(vi) आयरन(III)हेक्सासायनिडोफेरेट(II)

Answer»

OPTION no 2 is CORRECT.

Discussion

21.	निम्नलिखित संकुल स्पीशीज़ के चुंबकीय आघूर्णी के मान से आप क्या निष्कर्ष निकालेंगे?उदाहरण चुंबकीय आघूर्ण (BM) [tex] K4[Mn(CN)_6) 2.2[Fe(H_2O)6]^{2+} 5.3K_2[MnC_l4] 5.9[tex}
Answer» ANSWER: USE the FORMULA that's it the answer

21.

निम्नलिखित संकुल स्पीशीज़ के चुंबकीय आघूर्णी के मान से आप क्या निष्कर्ष निकालेंगे?उदाहरण चुंबकीय आघूर्ण (BM) [tex] K4[Mn(CN)_6) 2.2[Fe(H_2O)6]^{2+} 5.3K_2[MnC_l4] 5.9[tex}

Answer»

ANSWER:

USE the FORMULA that's it the answer

Discussion

22.	निम्नलिखित आयनों में प्रत्येक के लिए 3d इलेक्ट्रॉनों की संख्या लिखिए-आप इन जलयोजित आयनों (अष्टफलकीय) में पाँच 3d कक्षकों को किस प्रकार अधिग्रहीत करेंगे? दर्शाइए।
Answer» ANSWER: are U NEW over there

22.

निम्नलिखित आयनों में प्रत्येक के लिए 3d इलेक्ट्रॉनों की संख्या लिखिए-आप इन जलयोजित आयनों (अष्टफलकीय) में पाँच 3d कक्षकों को किस प्रकार अधिग्रहीत करेंगे? दर्शाइए।

Answer»

ANSWER:

are U NEW over there

Discussion

23.	निम्नलिखित उपसहसंयोजन यौगिकों के IUPAC नाम लिखिए-(i) [Co(NH3)6]Cl3 (iii) K3[Fe(CN)6] (v)K2[PdCl4] (ii)[Co(NH3)5Cl] Cl2 (iv) K3[Fe(C2O4)3] (vi) [Pt(NH3)2Cl (NH2CH3)]Cl
Answer» Answer: HI FRIEND how are you have a NICE day friend SORRY for the irreversible answer friend

23.

निम्नलिखित उपसहसंयोजन यौगिकों के IUPAC नाम लिखिए-(i) [Co(NH3)6]Cl3 (iii) K3[Fe(CN)6] (v)K2[PdCl4] (ii)[Co(NH3)5Cl] Cl2 (iv) K3[Fe(C2O4)3] (vi) [Pt(NH3)2Cl (NH2CH3)]Cl

Answer»

Answer:

HI FRIEND how are you have a NICE day friend SORRY for the irreversible answer friend

Discussion

24.	प्रथम संक्रमण श्रेणी के तत्व भारी संक्रमण तत्वों के अनेक गुणों से भिन्नता प्रदर्शित करते हैं। टिप्पणी कीजिए।
Answer» PLEASE mark me as BRAINLIEST. answer Is. . . . . . . . . I don't KNOW HINDI .

24.

प्रथम संक्रमण श्रेणी के तत्व भारी संक्रमण तत्वों के अनेक गुणों से भिन्नता प्रदर्शित करते हैं। टिप्पणी कीजिए।

Answer»

PLEASE mark me as BRAINLIEST.

answer Is. . . . . . . . .

I don't KNOW HINDI .

Discussion

25.	What coefficient must be placed in front of water in order for the equation to be balanced?CH4+2O2→CO2+H2O
Answer» ANSWER: here is your answer.......HOPE this HELP U....

25.

What coefficient must be placed in front of water in order for the equation to be balanced?CH4+2O2→CO2+H2O

Answer»

ANSWER:

here is your answer.......HOPE this HELP U....

Discussion

26.	Calculate the wave number of the light emitted corresponding to theFirst line of Balmer series in hydrogen spectrum
Answer» ν is WAVE number. For 4th LINE in Balmer series, initial and final STATES are 6 and 2, RESPECTIVELY.

Discussion

27.	5. Which of the following is correct [ ]a) Na > Na+ b) Cl – > Cl c) Both a and b d) Na < Na+
Answer» Answer: a) Na > Na+ Explanation: BECOZ when the E- is REMOVED the effective NUCLEAR force increases on REST of the e-

27.

5. Which of the following is correct [ ]a) Na > Na+ b) Cl – > Cl c) Both a and b d) Na < Na+

Answer»

Answer:

a) Na > Na+

Explanation:

BECOZ when the E- is REMOVED the effective NUCLEAR force increases on REST of the e-

Discussion

28.	Give 9th comparitive of 3 states of matter on the basis of rigidity
Answer» Answer: 3 states of MATTER are solid LIQUID and gas Explanation: on the basis of rigidity solid is the most RIGID the comes the liquid and then gas

28.

Give 9th comparitive of 3 states of matter on the basis of rigidity

Answer»

Answer:

3 states of MATTER are solid LIQUID and gas

Explanation:

on the basis of rigidity solid is the most RIGID the comes the liquid and then gas

Discussion

29.	A wire of about 2km length can bedrawn from one gram of Gold give reasons
Answer» Answer: Because gold is metal Metals have PROPERTY that they can be drawn into THIN wire so A wire of about 2KM LENGTH can be drawn from one gram of Gold Hope it helps you plzzzzzzzzz mark as brainlist PLEASE follow me and I will follow back

29.

A wire of about 2km length can bedrawn from one gram of Gold give reasons

Answer»

Answer:

Because gold is metal

Metals have PROPERTY that they can be

drawn into THIN wire so A wire of about 2KM LENGTH can be

drawn from one gram of Gold

Hope it helps you

plzzzzzzzzz mark as brainlist

PLEASE follow me and I will follow back

Discussion

30.	Olfactory indicators:
Answer» THANKS for FREE points It was the INDICATOR which indicate us by giving smell with ACIDS and BASES

30.

Olfactory indicators:

Answer»

THANKS for FREE points

It was the INDICATOR which indicate us by giving smell with ACIDS and BASES

Discussion

31.	Balance the equation
Answer» Answer: 2 NAOH + h2so4 = na2so4 + 2h2o 2al(oh)3 = AL2O3 +3H2O

31.

Balance the equation

Answer»

Answer:

2 NAOH + h2so4 = na2so4 + 2h2o

2al(oh)3 = AL2O3 +3H2O

Discussion

32.	Give reasons metals are used for making cooking vessels
Answer» Because it is HIGHLY mealleable DUCTILE and RESISTANCE of CORROSION

32.

Give reasons metals are used for making cooking vessels

Answer»

Because it is HIGHLY mealleable DUCTILE and RESISTANCE of CORROSION

Discussion

33.	Y cony are chemicals like lead nithate, Pottasiumchlorate and HgO stored away from sunlight
Answer» ANSWER: because it GETS DECOMPOSED

33.

Y cony are chemicals like lead nithate, Pottasiumchlorate and HgO stored away from sunlight

Answer»

ANSWER:

because it GETS DECOMPOSED

Discussion

34.	Atomic radius is measured by
Answer» Answer: The atomic radius (R) of an atom can be defined as ONE half the distance (d) between two nuclei in a diatomic molecule. Atomic radii have been measured for elements. The units for atomic radii are PICOMETERS, equal to 10−12 METERS. #HOPE IT HELPS........!!!!!!

34.

Atomic radius is measured by

Answer»

Answer:

The atomic radius (R) of an atom can be defined as ONE half the distance (d) between two nuclei in a diatomic molecule. Atomic radii have been measured for elements. The units for atomic radii are PICOMETERS, equal to 10−12 METERS.

#HOPE IT HELPS........!!!!!!

Discussion

35.	61, 91, 101 तथा 109 परमाणु क्रमांक वाले तत्वों का इलेक्ट्रॉनिक विन्यास लिखिए।
Answer» please MARK me as brainliest. ANSWER Is. . . . . . . I don't KNOW hindi .

35.

61, 91, 101 तथा 109 परमाणु क्रमांक वाले तत्वों का इलेक्ट्रॉनिक विन्यास लिखिए।

Answer»

please MARK me as brainliest.

ANSWER Is. . . . . . .

I don't KNOW hindi .

Discussion

36.	हुंड-नियम के आधार पर Ce3+ आयन के इलेक्ट्रॉनिक विन्यास को व्युत्पन्न कीजिए तथा ‘प्रचक्रण मात्र सूत्र' के आधार पर इसके चुंबकीय आघूर्ण की गणना कीजिए।
Answer» PLEASE MARK me as BRAINLIEST. answer Is. . . . . . . . . I don't know HINDI .

36.

हुंड-नियम के आधार पर Ce3+ आयन के इलेक्ट्रॉनिक विन्यास को व्युत्पन्न कीजिए तथा ‘प्रचक्रण मात्र सूत्र' के आधार पर इसके चुंबकीय आघूर्ण की गणना कीजिए।

Answer»

PLEASE MARK me as BRAINLIEST.

answer Is. . . . . . . . .

I don't know HINDI .

Discussion

37.	प्रथम श्रेणी के संक्रमण तत्वों के अभिलक्षणों को द्वितीय एवं तृतीय श्रेणी के वर्गों के संगत तत्वों से क्षैतिज वर्गों में तुलना कीजिए। निम्नलिखित बिंदुओं पर विशेष महत्व दीजिए-(i) इलेक्ट्रॉनिक विन्यास (ii) ऑक्सीकरण अवस्थाएं (iii) आयनन एन्थैल्पी तथा (iv) परमाण्वीय आकार
Answer» PLEASE mark me as BRAINLIEST. answer Is. . . . . . . . . I don't know HINDI .

37.

प्रथम श्रेणी के संक्रमण तत्वों के अभिलक्षणों को द्वितीय एवं तृतीय श्रेणी के वर्गों के संगत तत्वों से क्षैतिज वर्गों में तुलना कीजिए। निम्नलिखित बिंदुओं पर विशेष महत्व दीजिए-(i) इलेक्ट्रॉनिक विन्यास (ii) ऑक्सीकरण अवस्थाएं (iii) आयनन एन्थैल्पी तथा (iv) परमाण्वीय आकार

Answer»

PLEASE mark me as BRAINLIEST.

answer Is. . . . . . . . .

I don't know HINDI .

Discussion

38.	निम्नलिखित के संदर्भ में ऐक्टिनॉयड श्रेणी के तत्वों तथा लैन्थेनॉयड श्रेणी के तत्वों के रसायन की तुलना कीजिए। (i) इलेक्ट्रॉनिक विन्यास (ii) ऑक्सीकरण अवस्थाएं (iii) रासायनिक अभिक्रियाशीलता।
Answer» PLEASE MARK me as brainliest. ANSWER Is. . . . . . . . . I don't know HINDI . Answer Is ( i )

38.

निम्नलिखित के संदर्भ में ऐक्टिनॉयड श्रेणी के तत्वों तथा लैन्थेनॉयड श्रेणी के तत्वों के रसायन की तुलना कीजिए। (i) इलेक्ट्रॉनिक विन्यास (ii) ऑक्सीकरण अवस्थाएं (iii) रासायनिक अभिक्रियाशीलता।

Answer»

PLEASE MARK me as brainliest.

ANSWER Is. . . . . . . . .

I don't know HINDI .

Answer Is ( i )

Discussion

39.	लैन्थेनॉयड श्रेणी के उन सभी तत्वों का उल्लेख कीजिए जो +4 तथा जो +2 ऑक्सीकरण अवस्थाएं दर्शाते हैं। इस प्रकार के व्यवहार तथा उनके इलेक्ट्रॉनिक विन्यास के बीच संबंध स्थापित कीजिए।
Answer» Answer: आवर्त सारणी के पी ब्लॉक तत्वों के समूह 13,14,15,16,17 और 18 के होते हैं।इन तत्वों को भरने में सबसे बाहरी इलेक्ट्रॉनों के द्वारा विशेषता है p-अपने परमाणुओं के कक्षीय.इन तत्वों और उनके यौगिकों के कुछ एक महत्वपूर्ण खेलहमारे दैनिक जीवन में भूमिका। उदाहरण के लिए:नाइट्रोजन अमोनिया, नाइट्रिक अम्ल और उर्वरकों के निर्माण में प्रयोग किया जाता है। Trinitrotoluene(TNT), nitroglycrine, आदि, जो विस्फोटकों के रूप में उपयोग किया जाता यौगिकों के नाइट्रोजन, कर रहे हैं।हवा में मौजूद ऑक्सीजन जीवन और दहन प्रक्रियाओं के लिए आवश्यक हैकार्बोहाइड्रेट, प्रोटीन, विटामिन, एंजाइम, आदि, जो चेन कार्बन परमाणुओं के होते,विकास और रहने वाले जीव के विकास के लिए जिम्मेदार हैं।में सामान्य रुझान (अनुलंब रूप में अच्छी तरह के रूप में क्षैतिज) विभिन्न गुणों में मनायाएस-ब्लॉक भी इस ब्लाक में मनाया रहे हैं। के रूप में हम ऊपर से एक ऊर्ध्वाधर के माध्यम से नीचे की ओरस्तंभ (समूह) गुण में कुछ समानताएं मनाया जाता है। हालांकि, इस ऊर्ध्वाधरसमानता तुलना में विशेष रूप से एस-ब्लॉक, में मनाया कम पी ब्लॉक में चिह्नित किया गया है 13 और 15 समूहों; अनुलंब समानता तेजी से बाद में समूहों द्वारा दिखाया गया है। के रूप में दूर के रूप मेक्षैतिज प्रवृत्ति का संबंध है, हम भर में पंक्ति(period) करने के लिए दाईं ओर से बाईं ओर ले जाएँ के रूप में एक नियमित रूप से फैशन में गुण अलग-अलग। Explanation: PLEASE MARK ME AS BRAINLIEST and LIKE ME PLEASE

39.

लैन्थेनॉयड श्रेणी के उन सभी तत्वों का उल्लेख कीजिए जो +4 तथा जो +2 ऑक्सीकरण अवस्थाएं दर्शाते हैं। इस प्रकार के व्यवहार तथा उनके इलेक्ट्रॉनिक विन्यास के बीच संबंध स्थापित कीजिए।

Answer»

Answer:

आवर्त सारणी के पी ब्लॉक तत्वों के समूह 13,14,15,16,17 और 18 के होते हैं।इन तत्वों को भरने में सबसे बाहरी इलेक्ट्रॉनों के द्वारा विशेषता है p-अपने परमाणुओं के कक्षीय.इन तत्वों और उनके यौगिकों के कुछ एक महत्वपूर्ण खेलहमारे दैनिक जीवन में भूमिका। उदाहरण के लिए:नाइट्रोजन अमोनिया, नाइट्रिक अम्ल और उर्वरकों के निर्माण में प्रयोग किया जाता है। Trinitrotoluene(TNT), nitroglycrine, आदि, जो विस्फोटकों के रूप में उपयोग किया जाता यौगिकों के नाइट्रोजन, कर रहे हैं।हवा में मौजूद ऑक्सीजन जीवन और दहन प्रक्रियाओं के लिए आवश्यक हैकार्बोहाइड्रेट, प्रोटीन, विटामिन, एंजाइम, आदि, जो चेन कार्बन परमाणुओं के होते,विकास और रहने वाले जीव के विकास के लिए जिम्मेदार हैं।में सामान्य रुझान (अनुलंब रूप में अच्छी तरह के रूप में क्षैतिज) विभिन्न गुणों में मनायाएस-ब्लॉक भी इस ब्लाक में मनाया रहे हैं। के रूप में हम ऊपर से एक ऊर्ध्वाधर के माध्यम से नीचे की ओरस्तंभ (समूह) गुण में कुछ समानताएं मनाया जाता है। हालांकि, इस ऊर्ध्वाधरसमानता तुलना में विशेष रूप से एस-ब्लॉक, में मनाया कम पी ब्लॉक में चिह्नित किया गया है 13 और 15 समूहों; अनुलंब समानता तेजी से बाद में समूहों द्वारा दिखाया गया है। के रूप में दूर के रूप मेक्षैतिज प्रवृत्ति का संबंध है, हम भर में पंक्ति(period) करने के लिए दाईं ओर से बाईं ओर ले जाएँ के रूप में एक नियमित रूप से फैशन में गुण अलग-अलग।

Explanation:

PLEASE MARK ME AS BRAINLIEST and LIKE ME PLEASE

Discussion

40.	आंतरिक संक्रमण तत्व क्या हैं? बताइए कि निम्नलिखित में कौन से परमाणु क्रमांक आंतरिक संक्रमण तत्वों के हैं -29, 59, 74, 95, 102, 104
Answer» ANSWER: I don't knowZ!!!!!!!:-)!

40.

आंतरिक संक्रमण तत्व क्या हैं? बताइए कि निम्नलिखित में कौन से परमाणु क्रमांक आंतरिक संक्रमण तत्वों के हैं -29, 59, 74, 95, 102, 104

Answer»

ANSWER:

I don't knowZ!!!!!!!:-)!

Discussion

41.	ऐक्टिनॉयड तत्वों का रसायन उतना नियमित नहीं है जितना कि लैन्थेनॉयड तत्वों का रसायन। इन तत्वों की ऑक्सीकरण अवस्थाओं के आधार पर इस कथन का आधार प्रस्तुत कीजिए।
Answer» Answer: You realize EVERYONE on HEAR SPEAKS English or Spanish nooffence but we cant UNDERSTAND this Explanation:

41.

ऐक्टिनॉयड तत्वों का रसायन उतना नियमित नहीं है जितना कि लैन्थेनॉयड तत्वों का रसायन। इन तत्वों की ऑक्सीकरण अवस्थाओं के आधार पर इस कथन का आधार प्रस्तुत कीजिए।

Answer»

Answer:

You realize EVERYONE on HEAR SPEAKS English or Spanish nooffence but we cant UNDERSTAND this

Explanation:

Discussion

42.	A mixture of 0.5240 M CO and 0.4380 M Cl2 is enclosed in a vessel and heated to 1000 K .CO(g)+Cl2(g)↽−−⇀COCl2(g)c=255.0 at 1000 KCalculate the equilibrium concentration of each gas at 1000 K .
Answer» Answer: LET the initial velocity of the body be u m/s; and the constsnt acceleration be a m/s². Then displacemet (distance covered) of the body after t second is GIVEN by s = u t + ½ a t² ———————————-(1) The body covers 20 m in 2s and 80 m in 4 s. 20 = u×2 + ½ a 2²; => 20 = 2 u + 2 a u + a = 10 —————————————(2) 80 = u×4 + ½ a 4²; => 80 = 4 u + 8 a; this gives, u + 2 a = 20 ——————————————(3) Solving (2) and (3) for u and a we get u = 0 m/s and a = 10 m/s² ————————(4) To obtain the distance covered in the next 4s we need to find the velocity at the beginning of this interval using, V = u + a t => v = 0 m/s + 1.....

42.

A mixture of 0.5240 M CO and 0.4380 M Cl2 is enclosed in a vessel and heated to 1000 K .CO(g)+Cl2(g)↽−−⇀COCl2(g)c=255.0 at 1000 KCalculate the equilibrium concentration of each gas at 1000 K .

Answer»

Answer:

LET the initial velocity of the body be u m/s; and the constsnt acceleration be a m/s². Then displacemet (distance covered) of the body after t second is GIVEN by

s = u t + ½ a t² ———————————-(1)

The body covers 20 m in 2s and 80 m in 4 s.

20 = u×2 + ½ a 2²; => 20 = 2 u + 2 a

u + a = 10 —————————————(2)

80 = u×4 + ½ a 4²; => 80 = 4 u + 8 a; this gives,

u + 2 a = 20 ——————————————(3)

Solving (2) and (3) for u and a we get

u = 0 m/s and a = 10 m/s² ————————(4)

To obtain the distance covered in the next 4s we need to find the velocity at the beginning of this interval using,

V = u + a t => v = 0 m/s + 1.....

Discussion

43.	An atom contains 3 protons, 3 electrons, and 4 neutrons. Which of the following statements correctly describes the nucleus of this atom?A the nucleus contains 3 positively charged particlesB the nucleus contains 3 negatively charged particlesC the nucleus contains 3 positively charged particles and 4 neutrally charged particlesD the nucleus contains 3 negatively charged particles and 4 neutrally charged particlesGroup of answer choices
Answer» Answer: C. Atom have 3 proton , electron. and 4 neutron in NUCLEUS

43.

An atom contains 3 protons, 3 electrons, and 4 neutrons. Which of the following statements correctly describes the nucleus of this atom?A the nucleus contains 3 positively charged particlesB the nucleus contains 3 negatively charged particlesC the nucleus contains 3 positively charged particles and 4 neutrally charged particlesD the nucleus contains 3 negatively charged particles and 4 neutrally charged particlesGroup of answer choices

Answer»

Answer:

C. Atom have 3 proton , electron. and 4 neutron in NUCLEUS

Discussion

44.	Ch3chbrch2ch3+ oh- on heating gives
Answer» ANSWER: this is ELIMINATION reaction Explanation: ch2chch2ch3

44.

Ch3chbrch2ch3+ oh- on heating gives

Answer»

ANSWER:

this is ELIMINATION reaction

Explanation:

ch2chch2ch3

Discussion

45.	Describe an activity to show that particle of matter have spaces with diagram.among them. write the answer in steps
Answer» ANSWER: The space between the particles of matter can be shown by an activity of WATER and sugar. "When sugar is dissolved in the water, its crystals seperate into very fine particles. These particles of sugar go into the spaces between the various particles of water due to which there is no CHANGE in volume on DISSOLVING sugar in water. The fact that there is no change in volume on dissolving sugar in water tells us that these are spaces between the particles of water. And these spaces accomodate the sugar particles. This also gives us another conclusion that the particles(or molecules) in water are not TIGHTLY packed, they are somewhat loose, having spaces between them."

45.

Describe an activity to show that particle of matter have spaces with diagram.among them. write the answer in steps

Answer»

ANSWER:

The space between the particles of matter can be shown by an activity of WATER and sugar.

"When sugar is dissolved in the water, its crystals seperate into very fine particles. These particles of sugar go into the spaces between the various particles of water due to which there is no CHANGE in volume on DISSOLVING sugar in water. The fact that there is no change in volume on dissolving sugar in water tells us that these are spaces between the particles of water. And these spaces accomodate the sugar particles. This also gives us another conclusion that the particles(or molecules) in water are not TIGHTLY packed, they are somewhat loose, having spaces between them."

Discussion

46.	What is the valency of calcium ion
Answer» ANSWER: +2 is CORRECT answer Explanation: HOPE the answer will help you

46.

What is the valency of calcium ion

Answer»

ANSWER:

+2 is CORRECT answer

Explanation:

HOPE the answer will help you

Discussion

47.	Find the error in the following chemical equation and balance them. 4H2+O2=2H2O
Answer» EXPLANATION: products and REACTANTS must be equal in the above equation HYDROGENS are not balanced soo,2H2+O2------2H2O is correct

47.

Find the error in the following chemical equation and balance them. 4H2+O2=2H2O

Answer»

EXPLANATION:

products and REACTANTS must be equal

in the above equation HYDROGENS are not balanced

soo,2H2+O2------2H2O

is correct

Discussion

48.	Elements with genarally exhibit various oxidation states and form coloured ions are
Answer» question no ONE is GOING to be a GREAT DAY of SCHOOL and

Discussion

49.	The density of 3 molal sodium thiosulphate is 1.25 g/cm³. The mole fraction of sodium thiosulphate is
Answer» (i) Mass of 1000 mL of Na2S2O3 solution = 1.25×1000=1250g Mass of Na2S2O3 in 1000 mL of 3 M solution =3×MolmassofNa2S2O3 =3×158=474g Mass percentage of Na2S2O3 in solution =1250474×100=37.92 Alternatively, M=mAx×d×10 3=158x×1.25×10 x=37.92 (II) No. of moles of Na2S2O3=158474=3 Mass of WATER =(1250−474)=776g No. of moles of water =18776=43.1 Mole fraction of Na2

49.

The density of 3 molal sodium thiosulphate is 1.25 g/cm³. The mole fraction of sodium thiosulphate is

Answer»

(i) Mass of 1000 mL of Na2S2O3 solution = 1.25×1000=1250g

Mass of Na2S2O3 in 1000 mL of 3 M solution

=3×MolmassofNa2S2O3

=3×158=474g

Mass percentage of Na2S2O3 in solution

=1250474×100=37.92

Alternatively, M=mAx×d×10

3=158x×1.25×10

x=37.92

(II) No. of moles of Na2S2O3=158474=3

Mass of WATER =(1250−474)=776g

No. of moles of water =18776=43.1

Mole fraction of Na2

Discussion

50.	When 40.0 mL of 1.00 M H2SO4 is added to 80.0 mL of 1.00 M NaOH at 20.00°C in a coffee cup calorimeter, the temperature of the aqueous solution increases to 29.20°C. If the mass of the solution is 120.0 g and the specific heat of the calorimeter and solution is 4.184 J/g • °C, how much heat is given off in the reaction? (Ignore the mass of the calorimeter in the calculation.)Use q equals m C subscript p Delta T..A). 4.62 kJB). 10.0 kJC). 14.7 kJD). 38.5 kJ
Answer» Answer: A PARTICLE has a displacement of 40m along OX in the first 5s and 30M PERPENDICULAR to OX in the next 5s. The magnitude of average velocity of the particle during the time internal of 10s is a) 6 ms–1 B) 7 ms–1 c) 8 ms–1 d) 5 ms–1 [ Option d is the right answer.

50.

When 40.0 mL of 1.00 M H2SO4 is added to 80.0 mL of 1.00 M NaOH at 20.00°C in a coffee cup calorimeter, the temperature of the aqueous solution increases to 29.20°C. If the mass of the solution is 120.0 g and the specific heat of the calorimeter and solution is 4.184 J/g • °C, how much heat is given off in the reaction? (Ignore the mass of the calorimeter in the calculation.)Use q equals m C subscript p Delta T..A). 4.62 kJB). 10.0 kJC). 14.7 kJD). 38.5 kJ

Answer»

Answer:

A PARTICLE has a displacement of 40m along OX in the first 5s and 30M PERPENDICULAR to OX in

the next 5s. The magnitude of average velocity of the particle during the time internal of 10s is

a) 6 ms–1 B) 7 ms–1 c) 8 ms–1 d) 5 ms–1 [

Option d is the right answer.

Discussion

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Answer:

Explanation:

Therefore we get required answer.

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निम्नलिखित उपसहसंयोजन यौगिकों के IUPAC नाम लिखिए-(i) [Co(NH3)6]Cl3 (iii) K3[Fe(CN)6] (v)K2[PdCl4] (ii)[Co(NH3)5Cl] Cl2 (iv) K3[Fe(C2O4)3] (vi) [Pt(NH3)2Cl (NH2CH3)]Cl

प्रथम संक्रमण श्रेणी के तत्व भारी संक्रमण तत्वों के अनेक गुणों से भिन्नता प्रदर्शित करते हैं। टिप्पणी कीजिए।

What coefficient must be placed in front of water in order for the equation to be balanced?CH4+2O2→CO2+H2O

Calculate the wave number of the light emitted corresponding to theFirst line of Balmer series in hydrogen spectrum

5. Which of the following is correct [ ]a) Na > Na+ b) Cl – > Cl c) Both a and b d) Na < Na+

Give 9th comparitive of 3 states of matter on the basis of rigidity

A wire of about 2km length can bedrawn from one gram of Gold give reasons ​

Olfactory indicators:​

Balance the equation

Give reasons metals are used for making cooking vessels​

Y cony are chemicals like lead nithate, Pottasiumchlorate and HgO stored away from sunlight​

Atomic radius is measured by​

61, 91, 101 तथा 109 परमाणु क्रमांक वाले तत्वों का इलेक्ट्रॉनिक विन्यास लिखिए।

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ANSWER Is. . . . . . .

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A mixture of 0.5240 M CO and 0.4380 M Cl2 is enclosed in a vessel and heated to 1000 K .CO(g)+Cl2(g)↽−−⇀COCl2(g)c=255.0 at 1000 KCalculate the equilibrium concentration of each gas at 1000 K .

Ch3chbrch2ch3+ oh- on heating gives

Describe an activity to show that particle of matter have spaces with diagram.among them. write the answer in steps

What is the valency of calcium ion

Find the error in the following chemical equation and balance them. 4H2+O2=2H2O

Elements with genarally exhibit various oxidation states and form coloured ions are​

The density of 3 molal sodium thiosulphate is 1.25 g/cm³. The mole fraction of sodium thiosulphate is

1. Which of the following represents largest number of particles?a. atoms in 1 mole of CH4b. atoms in 0.5 mole of SO,c. atoms in 0.5 mole of CO4d. atoms in 1 mole of CO

(a) Giving reason identify the susanca ore and the substance reduced in each of the following chemical reactions :(i) H2 + CuO --> Cu + H2O(ii) H2S + SO2 --> 2S + 2H2O(b) Name the term used to represent a reaction in which ondation and reduction takepace simultaneously

Mass of total positive charge present in an atom is 16533 times to that of mass of electron.find the atomic number of this element.

Find density ofIceWaterfreonDiamondAluminiumWoodoilLeadIodine.

Calculate the amount of carbon dioxide produce in the combustion of 24gm of carbon

21+31=52 . write in scientific notion.....

Examples of slow and fast changes five of them

A wire of about 2km length can bedrawn from one gram of Gold give reasons

Olfactory indicators:

Give reasons metals are used for making cooking vessels

Y cony are chemicals like lead nithate, Pottasiumchlorate and HgO stored away from sunlight

Atomic radius is measured by

Elements with genarally exhibit various oxidation states and form coloured ions are